Question

The manager of a fleet of automobiles is testing two brands of radial tires and assigns one tire of each brand at random to the two rear wheels of eight cars and runs the cars until the tires wear out. The data (in kilometers) follow. Find a 99% confidence interval on the difference in the mean life.

Car Brand 1 Brand 2
1 37734 35202
2 45299 41635
3 36240 35500
4 32100 31950
5 37210 38015
6 48360 47800
7 38200 37810
8 33500 33215
(a) Calculate d=
(b) Calculate sD =
(c) Calculate a 99% two-sided confidence interval on the difference in mean life.

Answer 1

Brand 1 Brand 2 Difference

37734 35202 2532

45299 41635 3664

36240 35500 740

32100 31950 150

37210 38015 −805

48360 47800 560

38200 37810 390

33500 33215 285

Sum of difference = 2532+ 3664+740+150 −805+ 560 +390 +285 = 7516

Mean =
`d=(7516)/(8)`

Mean =
`d=939.5`

a) d= 939.5

`\text{Sample Standard deviation, s} = \sqrt{\frac{(x-\bar{x})^2}{n-1}}`

`=\sqrt{((2532-939.5)^2+(3664-939.5)^2+(740-939.5)^2 ...+(285-939.5)^2)/(8-1)}`

=1441.21

b)SD= 1441.21

c)Calculate a 99% two-sided confidence interval on the difference in mean life.

confidence level =99%

significance level =α= 0.01

Degree of freedom = n-1 = 8-1 =7

So,
`t_{(\alpha)/(2)}=3.499`

Formula for confidence interval
`= \left( \bar{X} \pm t_{(\alpha)/(2)} * (s)/(√(n)) \right)`

Substitute the values

confidence interval
`= 939.5 \pm 3.499 * (1441.21)/(√(8)) \right)`

confidence interval
`= 939.5 - 3.499 * (1441.21)/(√(8)) \right)` to
`= 939.5 + 3.499 * (1441.21)/(√(8)) \right)`

Confidence interval
`−843.396\` to
`2722.396`